#include <iostream> using namespace std; int baron(int h,int g){ cout<<h+g<<endl; cout<<h-g<<endl; cout<<h*g<<endl; cout<<h/g<<endl; cout<<h%g<<endl; } int main() { int h=4396,g=2200; baron(h,g); return 0; }

#include <iostream> using namespace std; int main() { int h=4396,g=2200; cout<<h+g<<endl; cout<<h-g<<endl; cout<<h*g<<endl; cout<<h/g<<endl; cout<<h%g<<endl; return 0; }

#include<bits/stdc++.h> using namespace std; int main() { int h=4396; int g=2200; cout<<h+g<<endl; cout<<h-g<<endl; cout<<h*g<<endl; cout<<h/g<<endl; cout<<h%g<<endl; return 0; }

A731.表达式计算 难度：入门 来源：官方 AC代码（C/C++）: 没有注释，自行学习。 by：ጿ ኈ ቼ ዽ ጿ ------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------ 你的点赞和关注就是我更新题解的最大动力！ ------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------ -----------------------------点个赞吧 ↓ ！-----------------------------

#include <iostream> using namespace std; int main() { int h=4396; int g=2200; cout<<h+g<<endl; cout<<h-g<<endl; cout<<h*g<<endl; cout<<h/g<<endl; cout<<h%g<<endl; return 0;

#include <iostream> using namespace std; int main() { int g=4396,h=2200; cout<<g+h<<endl; cout<<g-h<<endl; cout<<g*h<<endl; cout<<g/h<<endl; cout<<g%h<<endl; return 0; }

#（起因）突然想起Py #（经过）所以写代码 #（结果）然后就 AC #写work函数 def work(a, b): ____print(str(a+b)+'\n'+str(a-b)+'\n'+str(a*b)+'\n'+str(a//b)+'\n'+str(a%b)+'\n') work(4396,2200)#运行

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#include<bits/stdc++.h> using namespace std; int main(){ int H=4396,g=2200; cout<<H+g<<endl<<H-g<<endl<<H*g<<endl<<H/g<<endl<<H%g; return 0; }

#include <iostream> using namespace std; int main() { cout<<6596<<endl<<2196<<endl<<9671200<<endl<<1<<endl<<2196; }

#include <iostream> using namespace std; int main() { printf("6596\n2196\n9671200\n1\n2196\n"); return 0; }

using namespace std; int main( ) { int h=4396,g= 2200; cout <<h+g <<endl; cout <<h-g <<endl; cout<<h * g<<endl; cout<<h /g<<endl; cout<<h%g; }